I have the following question which is giving me a rather hard time.
Let $P$ be a $p$-subgroup of a finite group $G$. By considering an appropriate action of $P$, prove that $[G:P]$ is congruent to $[N_G(P):P]\bmod p$.
Hint: use the $p$ group fixed point theorem.
What I have tried so far. The $p$ group fixed point theorem states that if $G$ is a finite $p$-group acting on a finite set $X$. Let $X^G$ denote the subset of $X$ consisting of those elements fixed by $G$. Then $|X^G| \equiv |X| \bmod p$; in particular, if $p \nmid |X|$ then $G$ has a fixed point.
I tried to let $X$ be the set of all left cosets of $P$ in $G$. This way, $|X|=[G:P]$. However, I don't see a group action that would force $|X^G|=[N_G(P):P]$.
Let $P$ acts on $X=G/P$ by left multiplication. Then $$X^P=\{gP\mid xgP=gP\text{ for all }x\in P\}.$$ So, if $gP\in X^P$, we have $g^{-1}xg\in P$ for all $x\in P$. In other words, $g^{-1}Pg\subset P$. But both $P$ and $g^{-1}Pg$ have the same size, so they are in fact equal. Thus $g\in N_G(P)$. Similarly the converse also holds. Hence $X^P=N_G(P)/P$ as desired.